Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take a cyclohexane ring, put a positive charge on one ring carbon, and attach a methyl group to that same carbon. This is the one methyl cyclohexyl cation, and it has thirteen hydrogens in all. So how many hyperconjugation structures does it have? Is it all thirteen? Or is it only the three on the methyl group? Everybody has a guess, right? Hold on to yours; we will settle it in the last problem today. This is the JEE Main part of our organic chemistry problem bank: three quick problems first, and then two full ones. So, let's begin.

0:39Quick problem 1: count every bond, hydrogens too

Quick problem number one. Here is benzonitrile: C six H five, the phenyl group, joined to a C triple bond N group. The numbers written low count the atoms, the dash is a single bond, and the three short lines are a triple bond. How many sigma and pi bonds does it have? Sigma and pi are the Greek letters on screen. Okay? Let us count quickly: six bonds in the ring, one bond joining the ring to the C N carbon, and one sigma bond inside the triple bond. So eight sigma bonds, right? No, no, careful, that is exactly the trap! We forgot the five C H bonds, which this condensed formula does not draw as bonds. So use the course's rule: sigma bonds equal the number of atoms, minus one, plus the number of rings. Clear? The formula is C seven H five N: seven carbons, five hydrogens and one nitrogen, so thirteen atoms, and one ring. Sigma equals thirteen minus one, plus one, which gives thirteen sigma bonds. Right? Now the pi bonds. The ring has three double bonds, with one pi bond each, and the C triple bond N has two. Three plus two gives five pi bonds. So, thirteen sigma and five pi. With me?

1:55Quick problem 2: two chiral carbons, but meso?

Quick problem number two. How many stereoisomers does two bromo three chloro butane have, and is any of them meso? On screen is its structure: C H three, CHB r, CHC l, C H three. Bromine sits on carbon two, and chlorine on carbon three. Okay? Carbon two carries a hydrogen, a bromine, a methyl group and the rest of the chain: four different groups, so it is chiral. Carbon three is the same, with chlorine in place of bromine. So we have two chiral carbons. Got it? Now, two chiral carbons, side by side, in the middle of a butane chain. So one of the stereoisomers must be a meso form, right? No, no, wait, a meso form needs two identical halves! Here one half carries bromine and the other carries chlorine, so the halves are different. Then the count is two to the power n, where n is the number of chiral carbons; the n and the two written high are powers. So two squared gives four stereoisomers. Clear? With different halves, none of the four has a plane of symmetry. So there is no meso form, and all four are optically active. Okay?

3:06Quick problem 3: phenyl pulls, but distance weakens

Quick problem number three: arrange these in order of decreasing acidity. Ethanoic acid, C H three COOH; benzoic acid, with the COOH right on the phenyl ring; and phenylethanoic acid, where a C H two sits between the ring and the COOH group. With me? Now tell me. Compared with ethanoic acid, does the phenyl ring make the acid stronger, or weaker? It makes it stronger, very good! The phenyl group is minus I, where I stands for the inductive effect: it pulls electrons away from the carboxyl group, and groups that withdraw electrons increase acidity. Right? But inductive effects fall with distance. In phenylethanoic acid, the C H two keeps the ring one carbon away from the carboxyl group, so the pull is weaker. Okay? The p K a values agree, with the small a written low: four point seven six for ethanoic, four point one nine for benzoic, and four point three one for phenylethanoic acid. And remember, as the brackets say, the smaller the p K a, the stronger the acid. Clear? So the order is benzoic, greater than phenylethanoic, greater than ethanoic acid; the sign on screen means greater than. Got it?

4:21Example 1: sp3 carbons in 4-methylpent-1-yne

Alright, example one, an easy one. How many s p three carbons does four methyl pent one yne have? On screen, the three in s p three is written high. Pause the video and try it first. Done? Okay, let's check. Pent one yne is a five carbon chain with a triple bond starting at carbon one, and there is a methyl on carbon four. So it is H C triple bond C, then C H two, then C H with two methyl groups: the C H three in brackets, with a small two after it. One methyl is carbon five; the other is the branch. Okay? Now the course's one-line rule, for a carbon with no charge: count its pi bonds. No pi bond means s p three, one means s p two, and two means s p. Clear? Pent means five chain carbons, and two of them are in the triple bond. So three s p three carbons, right? Oh no, careful, we forgot the methyl branch! Go carbon by carbon. The two carbons of the triple bond each have two pi bonds, so both of them are s p. Okay? Every other carbon has only single bonds: the C H two, the C H, and both methyl groups. So all four of them are s p three. With me? So the answer is four s p three carbons, and two s p carbons. Always count the branches too. Got it?

5:45Example 2: hyperconjugation in 1-methylcyclohexyl cation

Now example two, the harder one, and it is our opening puzzle. How many hyperconjugation structures does the one methyl cyclohexyl cation have? Pause the video and try it first. Done? Okay, let's check, starting with the course figure. On the left is the ethyl cation, with the plus sign beside its positive carbon. Here, a C H sigma bond on the next carbon overlaps the empty p orbital, and the curved arrow shows those electrons moving in. Okay? On the right is propene, and the double-headed arrow links it to a no-bond resonance structure, where one hydrogen has left as H plus and the end carbon carries the minus sign. That H plus is not bonded to anything; the two hydrogens drawn above and below it belong to the carbon just to its right, which now has the double bond. Propene has three such structures, one for each alpha hydrogen, alpha being the Greek letter on screen. So the number of hyperconjugation structures equals the number of alpha hydrogens. With me? And which hydrogens are alpha? Only those on the carbons directly joined to the positive carbon, C plus. Right? Now our cation. The positive carbon is ring carbon one, and it also holds the methyl group. So which carbons are directly joined to it? Three of them, yes, exactly! The methyl carbon, with three hydrogens. And the two ring C H two groups on either side, at carbon two and carbon six, with two hydrogens each. Clear? So the total is three plus two plus two, which is seven. Not thirteen: the hydrogens at carbons three, four and five are too far away. And not three: the ring hydrogens next door count too. So that is our opening puzzle, completely solved.

7:38Quick revision: bond count, meso, -I effect, alpha H

So, quick revision. First, count every C H bond: sigma bonds equal atoms, minus one, plus rings. Second, when the two halves are different, there are two to the power n stereoisomers, and no meso form. Third, minus I groups raise acidity, and the effect falls with distance. And fourth, hyperconjugation structures equal the number of alpha hydrogens, on the carbons next door only. Now try the other problems on this page yourself, before reading their solutions. You are doing really well, so keep going, and all the best!